Coding Interview Questions

JOBHUNTER - "def findAlibaba(countOfRooms, strategy): #countofrooms: num rooms #listRooms rooms to look for alibabba possiblePlaces = [] #initialize rooms for i in range(countOfRooms): possiblePlaces.append(True) for i in range(len(strategy)): roomToCheck = strategy[i] #Room is marked as unavailable possiblePlaces[roomToCheck] = False #Next day calculatins nextDayPlaces = [] for j in range(countOfRooms): nextDayPla"See full answer

Casey C. - "Determine the requirements Perform basic checks on the frontend before submitting to the server Length Check for invalid or required characters Client-side validation messaging Perform more advanced checks on the server if required Does the password include the user's name, etc Handle server-side validation messaging"See full answer

Erjan G. - "this task is misleading . i used lag(1) and lead(1) cuz it did not say "compare temperature from 2 days before and 1 day before" , it reads to me as if its asking "compare cur temperature to prev and future and see if it rose and fall""See full answer

Jonathan michael J. - "add two strings `"See full answer

Bryan L. - "with var1 as (select *, rank() over(order by score desc) as srank from players) select player_name, score, srank as ranking from var1 where srank in (4, 6, 11) `"See full answer

Alvin P. - "SELECT e1.empid AS manageremployee_id, e1.empname AS managername, COUNT(e2.empid) AS numberofdirectreports FROM employees AS e1 INNER JOIN employees AS e2 ON e2.managerid = e1.empid GROUP BY e1.emp_id HAVING COUNT(e2.emp_id) >= 2 ORDER BY numberofdirectreports DESC, managername ASC `"See full answer

Alireza K. - "import random def coin_flip(): x=4*[0]+[1] res=[] for i in range(20): res.append(random.choice(x)) return res res=[0,0] # [head,tail] for j in range(1000): temp=coin_flip() res[0]+=sum(temp) #head res[1]+=(20-sum(temp)) #tail"See full answer

Ina K. - "HashMap supports insert, search, delete and retrieve in O(1). It stores data as key value pairs."See full answer

Jayveer S. - "SELECT p1.player_name AS player1, p2.player_name AS player2, ABS(p1.level - p2.level) AS level_disparity FROM players p1 JOIN players p2 ON p1.playername < p2.playername WHERE ABS(p1.level - p2.level) <= 5 ORDER BY level_disparity ASC;"See full answer

Derrick M. - "SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname FROM orders o INNER JOIN departments d ON d.departmentid = o.departmentid WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31' GROUP BY d.department_name; `"See full answer

Flora K. - "USING RECURSION"See full answer

Vince S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Nicholas S. - "Definitely nice to think of this without memorization, but there is a well known algorithm for this problem, which is the Levenshtein Distance. Lev(a,b) = len(a) if len(b) == 0 = len(b) if len(a) == 0 = lev(a[1:], b[1:] if a[0] == b[0] = 1 + min (lev(a, b[1:]), lev(a[1:], b), lev(a[1:], b[1:])) https://en.wikipedia.org/wiki/Levenshtein_distance I'm sure some optimizations could be made with heuristic."See full answer